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\newcommand{\Jaco}[2]{\frac{\partial (#1)} {\partial (#2)}}
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\pub{2009}{1}{3}{1}
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\topic{Lecture 12 \\Multiple Integral\\ \scriptsize Change of Variable (08 Oct 2009)}
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\section{Change of Variables} Let us consider the integral
\[\int\int_R f(x,y) dx dy\]
Variables appearing in this integral are $x,y$, Some times it becomes very useful to convert these variable to some other set of variables $u, v$ where $x,y$ and $u,v$ are connected by relations
\[x = \phi (u,v),~~~y=\psi (u,v)\]

If we use these relations, the integral $f(x,y)$ changes to some other function $F(u,v)$ of $u$ and $v$. Now question arises that what will happen with $dxdy$. Will these convert to $dudv$? No, not always. 

Let the equations $x = \phi (u,v),~~y=\psi (u,v)$ be solved for $u$ and $v$, so that
\[u = u(x,y), ~~ v=v(x,y)\]
The curves $u(x,y)=c_1$ and $v(x,y)=c_2$ form two systems of curves which devide the region $R$ in to elementary areas. Let $u, u+\delta u, v, v+\delta v$ be the curves enclosing one such area, say $PQRS$. In case of first order approximation the region $PQRS$ will be a parallelogram, and its areaa wouls be double that of the triangle $PQS$. If coordinates of $P$ are $(x,y)$ then coordinates of $Q,~S$ will be $\{x+\frac{\partial x}{\partial u} \delta u, y+\frac{\partial y}{\partial u} \delta u \}$ and $\{x+\frac{\partial x}{\partial v} \delta v, y+\frac{\partial y}{\partial v} \delta v \}$ respectively. Therefore the area of the element $PQRS$
\[\left|
\begin{array}{ccc}
x & y & 1 \\
x+\frac{\partial x}{\partial u} \delta u & y+\frac{\partial y}{\partial u} \delta u & 1 \\
x+\frac{\partial x}{\partial v} \delta v & y+\frac{\partial y}{\partial v} \delta v  & 1
\end{array}
\right|
\]

\[=\left|
\begin{array}{cc}
\frac{\partial x}{\partial u} \delta u & \frac{\partial y}{\partial u} \delta u  \\
\frac{\partial x}{\partial v} \delta v & \frac{\partial y}{\partial v} \delta v  
\end{array}
\right|
=
\left|
\begin{array}{cc}
\frac{\partial x}{\partial u}  & \frac{\partial y}{\partial u}   \\
\frac{\partial x}{\partial v}  & \frac{\partial y}{\partial v}   
\end{array}
\right|\delta u \delta v
=\Jaco{x,y}{u,v} \delta u \delta v
\]
Thus, if we take $x,y$  the elementary area is $\delta x \delta y$, which is in changing variable to $u,v$ becomes $\Jaco{x,y}{u,v} \delta u \delta v$. The limiting case will appear as 
\begin{equation}
dxdy = \Jaco{x,y}{u,v} dudv
\end{equation}
\begin{example}
Evaluate \[\int_0^2 \int_0^{\sqrt{2x-x^2}} (x^2+y^2) dydx\]
changing the variables to polar coordinates.
\end{example}

In polar coordinate system we have $x= r \cos \theta$, $y=r \sin \theta$.
Solve this problem in three step : First compute new limits, second change variables $x$, $y$ and the elementary area $dxdy$, Then in third step solve it.

$0 \leq x \leq 2 $, represents the region between line $x=0$ and the line $x=2$, and $0 \leq y \leq \sqrt{2x-x^2} $ represents the region bounded below by line $y=0 $ and upper by the circle $y=\sqrt{2x-x^2} $ or $x^2+y^2 - 2x = 0$. This shows that the region is the upper half circle, which has its diameter on x axis and touches origin. Let us convert this equation of circle in polar form, put $x= r \cos \theta$, $y=r \sin \theta$, we get 
\[r^2 - 2r \cos \theta =0\]
which gives limits of $r$ are $0$ to $2 \cos \theta$. Limits of $\theta$ are $0$ to $\pi / 2$ as whole upper first quadrant is in region.

Now look at integrand - quantity $x^2+y^2$ becomes $r^2$ and $dxdy$ becomes $rdrd\theta$ as $\Jaco{x,y}{r,\theta}=r$. Thus integral becomes
\[\int_0^{\pi/2} \int_0^{2 \cos \theta} r^3 drd\theta = \frac{3 \pi}{4}\]
\section*{Problems}
\begin{enumerate}
\item  Transform $\int _{0}^{\pi /2} $$\int _{0}^{\pi /2} $$\sqrt{\left(\frac{\sin \phi }{\sin \theta } \right)} $d $\phi $ d $\theta ,$by the substitution x = sin $\phi \sin \theta ,$and show that its value is  $\pi $
\item  Using the transformation $x+y=u$, $y = uv$, show that
\[\int \int \left[xy(1-x-y)\right]^{1/2}dx dy = \frac{2\pi }{105}\]
integration being taken over the area of the triangle bounded by the lines $x = 0$, $y = 0$, $x + y =1$.
\item  Evaluate $\int _{0}^{2} \int _{0}^{\sqrt{2x-x^{2} } } (x^{2} +y^{2} ) dy dx$.
\item  Evaluate $\int _{0}^{\infty }\int _{0}^{\infty } e^{-(x^2+y^2)} dxdy$ by changing to polar co-ordinates. Hence, show that $\int _{0}^{\infty } e^{-x^2} dx =\frac{\sqrt{\pi } }{2} $.
\end{enumerate}

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